A long solenoid with 30.0 turns/cmand a radius of 8.50 cm carries acurrent of 14

Question

A long solenoid with 30.0 turns/cmand a radius of 8.50 cm carries acurrent of 14.0 mA. A current of7.00 A exists in a straight conductorlocated along the central axis of the solenoid. (a) At what radial distance from the axis willthe direction of the resulting magnetic field be at 45.0° tothe axial direction?
1 cm

(b) What is the magnitude of the magnetic field there?
2 T















(a) At what radial distance from the axis willthe direction of the resulting magnetic field be at 45.0° tothe axial direction?
1 cm

(b) What is the magnitude of the magnetic field there?
2 T

Explanation / Answer

given that length of solinoid = L =30.0 turns/cm radius = 8.50cm = 8.5 * 10-2m current = I = 14.0mA = 14 * 10-3A magnetic field at point P on the axis due tosolenoid and the wire as Bs andBw so Bs is perpendicular toBw net electric field B to be at450 with the axis when Bs =Bw Bs = 0 is n =Bw = 0 iw/2d    from the above eq d = iw/2is n     = 7.00A/2 (14 * 10-3 A) (30 turns/cm) solve d = -------cm the magnetic field stength = B = 2 Bs                                               = 2(4 * 10-7 T.m/A ( 14 * 10-3A( 30 turns/0.01m ) solve B = --------T                                               = 2(4 * 10-7 T.m/A ( 14 * 10-3A( 30 turns/0.01m ) solve B = --------T

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